91Ó°ÊÓ

1. Flywheel rotation is $25\text{rad}/\text{sec}$
2. Flywheel stops in $20\text{sec}$

Here, initial and final angular velocity and time are given. Using the angular kinematic equations, find angular acceleration and angle. Convert the angle into revolutions by using the relationship between the radians traced in one rotation.

Formulae are as follow:

$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }\text{t}$

$\mathrm{θ}=\frac{\mathrm{Ó\neg }+{\mathrm{Ó\neg }}_0}{2}×\text{t}$

Where,

t is time, $\mathrm{Ó\neg },  {\mathrm{Ó\neg }}_0$ are final and initial angular velocities, is angular acceleration and $\mathrm{θ}$ is displacement.

Using the angular kinematic equation,

$$\begin{gathered} \mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }\text{t} \\ 0=25+\mathrm{Î\pm }×20 \\ \mathrm{Î\pm }=-1.25\text{rad}/s^2 \end{gathered}$$

Hence, angular acceleration of the flywheel is $\text{- 1.25}\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$

The angular displacement can be found as,

$$\begin{gathered} \mathrm{θ}=\left(\frac{\mathrm{Ó\neg }+{\mathrm{Ó\neg }}_0}{2}\right)\text{t} \\ \mathrm{θ}=\frac{25}{2}×2 \\ \mathrm{θ}=250\text{rad} \end{gathered}$$

Hence, angle through which the flywheel rotates is $250\text{rad}$

Convertangular displacement into number of rotations as,

$$\begin{gathered} \text{n}=250\text{rad}×\frac{1\text{rev}}{2×\mathrm{π}} \\ \text{n}=39.8\text{rev} \end{gathered}$$

Hence, number of revolutions made by the flywheel is$39.8\text{rev}$.

Therefore, basic formulas of rotational kinematics can be used to find angular acceleration, angle, and number of revolutions.